200 mL of a strong acid solution of pH 2.0 is mixed with 800 mL of another acid solution of pH 3.0. The pH of the resultant solution is

Options

A. `2.55`
B. `2.97`
C. `2.40`
D. `2.10`

Explanation

To solve the problem of finding the pH of the resultant solution when mixing two acid solutions, we will follow these steps:

### Step 1: Calculate the concentration of H⁺ ions in both solutions.

1. **For the first solution (pH 2.0)**:
- pH = 2.0
- \[ [H^+]_1 = 10^{-pH} = 10^{-2} \, \text{M} = 0.01 \, \text{M} \]

2. **For the second solution (pH 3.0)**:
- pH = 3.0
- \[ [H^+]_2 = 10^{-pH} = 10^{-3} \, \text{M} = 0.001 \, \text{M} \]

### Step 2: Calculate the number of moles of H⁺ ions in each solution.

1. **For the first solution (200 mL)**:
- Volume = 200 mL = 0.200 L
- Moles of H⁺ from solution 1:
\[ n_1 = [H^+]_1 \times \text{Volume} = 0.01 \, \text{M} \times 0.200 \, \text{L} = 0.002 \, \text{moles} \]

2. **For the second solution (800 mL)**:
- Volume = 800 mL = 0.800 L
- Moles of H⁺ from solution 2:
\[ n_2 = [H^+]_2 \times \text{Volume} = 0.001 \, \text{M} \times 0.800 \, \text{L} = 0.0008 \, \text{moles} \]

### Step 3: Calculate the total moles of H⁺ ions in the mixed solution.

- Total moles of H⁺:
\[ n_{total} = n_1 + n_2 = 0.002 + 0.0008 = 0.0028 \, \text{moles} \]

### Step 4: Calculate the total volume of the mixed solution.

- Total volume:
\[ V_{total} = 200 \, \text{mL} + 800 \, \text{mL} = 1000 \, \text{mL} = 1.0 \, \text{L} \]

### Step 5: Calculate the concentration of H⁺ ions in the resultant solution.

- Concentration of H⁺ in the resultant solution:
\[ [H^+]_{resultant} = \frac{n_{total}}{V_{total}} = \frac{0.0028 \, \text{moles}}{1.0 \, \text{L}} = 0.0028 \, \text{M} \]

### Step 6: Calculate the pH of the resultant solution.

- pH of the resultant solution:
\[ pH = -\log[H^+]_{resultant} = -\log(0.0028) \]

Using a calculator:
\[ pH \approx 2.55 \]

### Final Answer:
The pH of the resultant solution is approximately **2.55**.

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